Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $q = \dfrac{5k^2 - 30k}{k^3 - 5k^2 - 6k} \times \dfrac{-k^2 + 5k + 6}{k + 5} $
Explanation: First factor out any common factors. $q = \dfrac{5k(k - 6)}{k(k^2 - 5k - 6)} \times \dfrac{-(k^2 - 5k - 6)}{k + 5} $ Then factor the quadratic expressions. $q = \dfrac {5k(k - 6)} {k(k + 1)(k - 6)} \times \dfrac {-(k + 1)(k - 6)} {k + 5} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac {5k(k - 6) \times -(k + 1)(k - 6) } { k(k + 1)(k - 6) \times (k + 5)} $ $q = \dfrac {-5k(k + 1)(k - 6)(k - 6)} {k(k + 1)(k - 6)(k + 5)} $ Notice that $(k + 1)$ and $(k - 6)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac {-5k\cancel{(k + 1)}(k - 6)(k - 6)} {k\cancel{(k + 1)}(k - 6)(k + 5)} $ We are dividing by $k + 1$ , so $k + 1 \neq 0$ Therefore, $k \neq -1$ $q = \dfrac {-5k\cancel{(k + 1)}(k - 6)\cancel{(k - 6)}} {k\cancel{(k + 1)}\cancel{(k - 6)}(k + 5)} $ We are dividing by $k - 6$ , so $k - 6 \neq 0$ Therefore, $k \neq 6$ $q = \dfrac {-5k(k - 6)} {k(k + 5)} $ $ q = \dfrac{-5(k - 6)}{k + 5}; k \neq -1; k \neq 6 $